Calculating Moon Azimuth

Hello,

If I have a point on earth with with specific Lat and Long. And if I
have the Lat and Long of the moon for specific time. How can I get the
Azimuth between the point on earth and the moon? I'm trying to draw the
earth in circle and show the moon location with respect to the point on
earth for a specific time. Is the Azimuth what i really need?

N
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W ------- E
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| O moon
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I'm not expert on this subject but I have program that gives me the
current lat and long of the moon at specific time but I do not know how
to convert this location to degrees where i can plot it with respect to
the observer lat and long. Any formulas or examples is greatly
appreciated.


Thanks for your help

B M wrote:
If I have a point on earth with with specific Lat and Long. And if I
have the Lat and Long of the moon for specific time.

Are you sure you have the latitude and longitude of the Moon? Or do you
instead have the right ascension (RA) and declination (Dec) of the Moon?

--
Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
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According to the source code, I think it is the ecliptic lat and
ecliptic long? Would that be the same as Lat and Long?

B M wrote:
According to the source code, I think it is the ecliptic lat and
ecliptic long? Would that be the same as Lat and Long?

No. Latitude and longitude on the Earth is reckoned from the Earth's
axis and the zero longitude line through Greenwich; ecliptic latitude
and longitude are reckoned from the ecliptic axis (perpendicular to the
Earth's orbit) and the zero point of Aries. The two axes are separated
by the 23.4 degree angle of the Earth's tilt, and what's more, the
Earth's latitude/longitude system is constant rotating with respect to
the ecliptic longitude/latitude system, at one revolution per sidereal
day.

Whew. After all that, what you need to do is to convert ecliptic
latitude and longitude into equatorial coordinates. I think Paul
Schlyter's page has some tips on how to do that.

--
Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html

"B M" wrote in message oups.com...
According to the source code, I think it is the ecliptic lat and
ecliptic long? Would that be the same as Lat and Long?

No.

The ecliptic is the apparent path the sun traces
around the sky over the course of a year. Latitude
and longitude are coordinates relative to the Earth's
equator and axis. Like the equator the ecliptic is a
circle with its center at the center of the Earth, however
the ecliptic is tilted relative to the equator.

It sounds like what you want to do is compute the Moon's
altitude and azimuth given its Right Ascension(RA) and
Declination(DEC). This is called an equatorial to horizon
coordinate transformation. If you can get a copy of
Practical Astronomy with your Calculator, 3rd Edition
by Peter Duffett-Smith, he explains shows how to do this
conversion early in the book.

A website where you can get lunar coordinates is http://users.zoominternet.net/~matto...0ephemeris.htm

The free java program Nightvision http://home.att.net/~bsimpson/nvj.html
calculates the Moon's position and also gives altitude and azimuth.

Hilton Evans
---------------------------------------------------------------
Lon -71° 04' 35.3"
Lat +42° 11' 06.7"
---------------------------------------------------------------
Webcam Astroimaging
http://mysite.verizon.net/hiltonevan...troimaging.htm
---------------------------------------------------------------
ChemPen Chemical Structure Software
http://www.chempensoftware.com

B M wrote:
If I have a point on earth with with specific Lat and Long. And if I
have the Lat and Long of the moon for specific time. How can I get the
Azimuth between the point on earth and the moon? . . . I'm trying to draw the
earth in circle and show the moon location with respect to the point on
earth for a specific time. Is the Azimuth what i really need?

For the first part of your question regarding an Earth-Moon diagram and
using simple graphical methods, you can also just plot a simple diagram
of the position of the Moon over a series of days using the Moon's
ecliptic coordinates using a protractor and the Moon's distance.

Simply draw the Earth circle and establish a fundamental line anywhere
on the disk. Use the protractor to mark off the degrees of ecliptic
longitude from the fundamental line. The establish a scale for the
Moon's distance, e.g. 12cm = 400,000 km. Mark the Moon's distance
using this scale.

For the first part of your question to plot the position of the Moon
and using computational instead of graphical methods, you do not need
to convert to your local horizon system coordinates of altitude and
azimuth to draw your Earth-Moon circle diagram. If you have the Moon's
coordinates in the ecliptic coordinate system, you only need the the
distance to the Moon.

Then you need to convert ecliptic coordinates into x,y,z Earth-centered
(or orthogonal) coordinates.

First, become familiar with an online web applications that will return
the Moon's ecliptic coordinates and/or your local horizon system
altitude and azimuth and the Moon's current distance - the NASA/JPL
Horizons web applet:
NASA/JPL Ephemeris Generator http://ssd.jpl.nasa.gov/horizons.cgi

The Horizons web interface is a little klunky to use, but once you get
the hang of it, it is the "gold-standard" of online ephemeris
applications

Now with the Moon's distance and ecliptic coordinates, convert to
orthogonal coordinates.

Here's my partial VBA code snippet for converting the ecliptic
coordinates and distance to x,y,z coordinates:

Dim gamma As Double ' Moon ecliptic longitude
Dim beta As Double ' Moon ecliptic latitude
Dim eps As Double ' obliquity of the ecliptic
Dim R_km As Double ' Distance from the Moon in kilometers geocentric
Dim s_x_km, s_y_km, s_z_km As Double ' equatorial (geocentric)
rectangular coordinates of the Moon in kilometers

' Convert s_x, s_y, s_z, the rectangular coordinates of the Moon in
kilometers
s_x_km = R_km * Cos(Deg2Rad(gamma)) * Cos(Deg2Rad(beta))
' Clear working variables
A = 0
B = 0

' Find y coord
A = Cos(Deg2Rad(eps)) * Sin(Deg2Rad(gamma)) * Cos(Deg2Rad(beta))
B = Sin(Deg2Rad(beta)) * Sin(Deg2Rad(eps))
s_y_km = R_km * (A - B)
' Clear working variables
A = 0
B = 0

' Find z coord
A = Sin(Deg2Rad(eps)) * Sin(Deg2Rad(gamma)) * Cos(Deg2Rad(beta))
B = Sin(Deg2Rad(beta)) * Cos(Deg2Rad(eps))
s_z_km = R_km * (A + B)
' Clear working variables
A = 0
B = 0

Sorry, I don't have the specific source on the basic trig equations (at
my current physical location) that I used to convert to celestial or
ecliptic to x,y,z. (I'll look them up later and post an update.) Maybe
Brian Tung has some handy. This web page may be of help:
http://www.colorado.edu/geography/gc...oordsys_f.html


The lead books with examples of how the computations are performed a


Duffet-Smith1988: Duffet-Smith, P. 1988 (3ed). Practical Astronomy with
Your Calculator. Cambridge Press.
1988QB62.5.D83..... http://adsabs.harvard.edu/cgi-bin/np...1988QB62.5.D83.....

Montenbruck, O. & Pfleger, T. 2002. 4ed. Astronomy on the Personal
Computer. Springer. ISBN 3-540-67221-4 http://www.springer.com/

Meeus, J. 1998. 2ed. Astonomical Algorithms. Willmann-Bell. ISBN
0-943396-61-1 http://www.willbell.com/

(I'd recommend going to your local college library and copying the
applicable sections out of Montenbruck and Duffet-Smith.)

I haven't gone into the second part of your question - what is the
position of the observer on the Earth's globe in your diagram. Please
digest this first part. Maybe that will flush out your question
better.

- Canopus56

P.S. For some online general coordiante calculators (including exposed
javascript code):

Schmitt's online Celestial to Horizon Calculator
http://home.att.net/~srschmitt/scrip...l2horizon.html

Togo's online coordinate calculator
F:\Daily\Observing tools\Coordinate conversion Roman java.htm

If you want to dive deep in this area, take a look at some of the books
in the Math and Celestial Mechanics section at Willman-Bell - the lead
publisher in this area:
http://www.willbell.com/

Also in the Almanacs section
http://www.willbell.com/almanacs/index.htm , consider the MICA 2.0
Interactive Almanac by the U.S. Naval Obs.

Thanks for all the posts!!! This really help, i will try it and let you
know.

Canopus56 please if you find that code can you post it?

Thanks again.

Brian Tung wrote:
B M wrote:
According to the source code, I think it is the ecliptic lat and
ecliptic long? Would that be the same as Lat and Long?

No. Latitude and longitude on the Earth is reckoned from the Earth's
axis and the zero longitude line through Greenwich; ecliptic latitude
and longitude are reckoned from the ecliptic axis (perpendicular to the
Earth's orbit) and the zero point of Aries. The two axes are separated
by the 23.4 degree angle of the Earth's tilt, and what's more, the
Earth's latitude/longitude system is constant rotating with respect to
the ecliptic longitude/latitude system, at one revolution per sidereal
day.


To add further to your freakish description,the sidereal consequence
is that the Earth maintains a constant face to the Sun when it simply
does not -

http://www.pfm.howard.edu/astronomy/...S/AACHCIR0.JPG

In an era where it is vital to get accurate working principles for
axial/orbital motions and orientations for climatology ,this is how you
look on things !.

The Earth orbital orientation,assigned through the division betwen
direct Sunlight and the orbital shadow changes over the course of an
annual orbit causing the shadow to move across either poles at specific
points in the Earths' orbit.

The extreme fixed tilt of Uranus demonstrates that orbital orientation
changes and it is no different for the Earth

http://physics.uoregon.edu/~jimbrau/...13/FG13_06.jpg

Your celestial sphere and the sidereal justification of it is creating
havoc where an accurate version of the Earths' motions and orientations
are required.The horrendous thing is that I remain the sole voice
presently in the midst of this intellectual and intuitive holocaust and
it is time for others to actively recover the lost working astronomical
blueprints of Copernicus and Kepler.






Whew. After all that, what you need to do is to convert ecliptic
latitude and longitude into equatorial coordinates. I think Paul
Schlyter's page has some tips on how to do that.

--
Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html